A solid spherical ball rolls on a horizontal | vedclass

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(C) Given:Initial velocity of the solid spherical ball is $v = 10 \ m \ s^{-1}$.Mass of the ball is $m = 11 \ kg$.Since frictional losses are negligib

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$7$

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(C) Given:Initial velocity of the solid spherical ball is $v = 10 \ m \ s^{-1}$.Mass of the ball is $m = 11 \ kg$.Since frictional losses are negligible, the total mechanical energy is conserved.The initial total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energy:$K_i = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$For a solid sphere, the moment of inertia $I = \frac{2}{5} m R^2$ and $\omega = \frac{v}{R}$.$K_i = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2 = \frac{1}{2} m v^2 (1 + \frac{2}{5}) = \frac{1}{2} m v^2 (\frac{7}{5}) = \frac{7}{10} m v^2$.At the maximum height $h$, the ball momentarily stops, so its final kinetic energy $K_f = 0$.The potential energy gained by the ball is $U_f = m g h$.By the law of conservation of energy, $K_i = U_f$:$\frac{7}{10} m v^2 = m g h$$h = \frac{7 v^2}{10 g} = \frac{7 	imes (10)^2}{10 	imes 10} = \frac{7 	imes 100}{100} = 7 \ m$.Thus, the value of $h$ is $7 \ m$.

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